Question 183194
{{{x^2+y^2-10x+4y+13=0}}} Start with the given equation.



{{{x^2+y^2-10x+4y=-13}}} Subtract 13 from both sides.



{{{(x^2-10x)+(y^2+4y)=-13}}} Group like terms.



{{{(x^2-10x+highlight(25))+(y^2+4y)=-13+highlight(25)}}} Take half of the "x" coefficient -10 to get -5. Square it to get 25. Add this to both sides.



{{{(x^2-10x+25)+(y^2+4y+highlight(4))=-13+25+highlight(4)}}} Take half of the "y" coefficient 4 to get 2. Square that result to get 4. Add this to both sides.



{{{(x^2-10x+25)+(y^2+4y+4)=-13+25+4}}} 



{{{(x^2-10x+25)+(y^2+4y+4)=16}}} Combine like terms.



{{{(x-5)^2+(y^2+4y+4)=16}}} Factor {{{x^2-10x+25}}} to get {{{(x-5)^2}}}



{{{(x-5)^2+(y+2)^2=16}}} Factor {{{y^2+4y+4}}} to get {{{(y+2)^2}}}



{{{(x-5)^2+(y+2)^2=4^2}}} Rewrite {{{16}}} as {{{4^2}}}



{{{(x-5)^2+(y-(-2))^2=4^2}}} Rewrite {{{y+2}}} as {{{y-(-2)}}}



So the equation is now in the form {{{(x-h)^2+(y-k)^2=r^2}}} (which is a circle) where (h,k) is the center and "r" is the radius



We can see that {{{h=5}}}, {{{k=-2}}}, and {{{r=4}}}



So the center is (5,-2) and the radius is 4 units